Worked example: Approximation with local linearity | AP Calculus AB | Khan Academy

We're told the function ( f ) is twice differentiable with ( f(2) = 1 ), ( f'(2) = 4 ), and ( f''(2) = 3 ). What is the value of the approximation of ( f(1.9) ) using the line tangent to the graph of ( f ) at ( x = 2 )? So pause this video and see if you can figure this out. This is an actual question from a past AP calculus exam.

All right, now let's do this together. If I was actually doing this on the exam, I would just cut to the chase and I would figure out the equation of the tangent line at ( x = 2 ) going through the point ( (2, 1) ), and then I would figure out, okay, when ( x = 1.9 ), what is the value of ( y )? That would be my approximation. But for the sake of learning and getting the intuition here, let's just make sure we understand what's happening.

So let me graph this. Let's say that's my ( y )-axis, and then this is my ( x )-axis. This is ( x = 1 ), this is ( x = 2 ), this is ( y = 1 ). We know that the point ( (2, 1) ) is on the graph of ( y = f(x) ), so we know that point right over there is there. And we also know the slope of the tangent line. The slope of the tangent line is ( 4 ). So it's going to look something like this; it's going to probably even be a little steeper than that.

The slope of the tangent line is going to look something like that. We don't know much more about it; we know the second derivative here. But what they're asking us to do is, without knowing what the function actually looks like, the function might look something like this. Let me just draw something. So the function might look something like this.

We're trying to figure out what ( f(1.9) ) is, so if ( x = 1.9 ), ( f(1.9) ) — if that's the way the function actually looked — might be this value right over here. But we don't know for sure because we don't know much more about the function. What they're suggesting for us to do is use this tangent line.

If we know the equation of this tangent line here, we can say, well, what does that tangent line equal when ( x = 1.9 )? When ( x = 1.9 ), it equals that point right over there, and then we could use that as our approximation for ( f(1.9) ).

Well, to do that, we know we need to know the equation of the tangent line, and we could do that in point-slope form. We would just have to say ( y - ) the ( y ) value that we know is on that line. The point ( (2, 1) ) we know is on that line, so ( y - 1 ) is going to be equal to the slope of our tangent line, which we know is going to be equal to ( 4 ) times ( x - ) the ( x ) value that corresponds to that ( y ) value, so ( x - 2 ).

So now we just have to substitute ( x = 1.9 ) to get our approximation for ( f(1.9) ). So we'd say ( y - 1 = 4(1.9 - 2) ). ( 1.9 - 2 ) is ( -0.1 ), and let's see, ( 4 \times -0.1 ) — this all simplifies to ( -0.4 ).

Now you add ( 1 ) to both sides; you get ( y = 1 - 0.4 ). If you add ( 1 ) here, you're gonna get ( 0.6 ). So this — I didn't draw it quite to scale — ( 0.6 ) might be something closer to right around there, but there you go. That is our approximation for ( f(1.9) ), which is choice ( b ), and we're done.

One interesting thing to note is we didn't have to use all the information they gave us. We did not have to use this information about the second derivative in order to solve the problem. So if you ever find yourself in that situation, don't doubt yourself too much because they will sometimes give you unneeded information.